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If I'm dealing with extrinsic Euler angles are the Euler rates equal to the angular rates? This seems logical to me because there aren't any intermediate frames; however, I'm instead only seeing this formula for body rates

$$ \left( \begin{array}{c} p \\ q \\ r \\ \end{array} \right) = \left( \begin{array}{ccc} 1 & 0 & -\sin{\theta} \\ 0 & \cos{\phi} & \sin{\phi} \cos{\theta} \\ 0 & -\sin{\phi} & \cos{\phi} \cos{\theta} \\ \end{array} \right) \left( \begin{array}{c} \dot{\phi} \\ \dot{\theta} \\ \dot{\psi} \\ \end{array} \right) $$

But it seems to me that for extrinsic rotations $$ \left( \begin{array}{c} \omega_x \\ \omega_y \\ \omega_z \\ \end{array} \right) = \left( \begin{array}{c} \dot{\phi} \\ \dot{\theta} \\ \dot{\psi} \\ \end{array} \right) $$

and the body rates are the angular rates rotated into the body frame (via some rotation matrix $\mathbf{T}$)

$$ \left( \begin{array}{c} p \\ q \\ r \\ \end{array} \right) = \mathbf{T} \left( \begin{array}{c} \omega_x \\ \omega_y \\ \omega_z \\ \end{array} \right) $$

Am I incorrect? Or missing something?

user1543042
  • 486

1 Answers1

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I looked into this more and you can derive the relationship between the rotational rates and Euler rates using the expression

$$\Omega = -\dot{\mathbf{T}} \mathbf{T}^{\mathrm{T}}$$

where $\Omega$ is the skew-symmetric matrix of the rotational rate $\omega$

$$\Omega = \left( \begin{array}{ccc} 0 & -\omega_z & \omega_y \\ \omega_z & 0 & -\omega_x \\ -\omega_y & \omega_x & 0 \\ \end{array} \right) $$

This is a completely general expression, it comes from the fact that

$$\mathbf{T} \mathbf{T}^{\mathrm{T}}=\mathbf{1}$$

See a good tutorial here

user1543042
  • 486